Spinning liquid creating a parabolic shape

take a container of liquid and spin it... a parcel at any given radius (r) experiences (in the rotating frame) a force outward = mass x angular speed^2 x r and a vertical force of mg. Thus the surface curves such that the tangent to the surface makes an angle to the horizontal given by tan theta = angular speed^2 x r/g.

two questions:
-I think this is simple and I'm just missing it... how do I prove that the resultant shape is a parabola?

-what provides the force we see at work? surface tension? pressure gradient? Basically how do we draw a free body diagram for the parcel of liquid in the Earth frame?
Thank you!

It's simply that all the liquid wants to get to the outer edge of the bucket.
Since there is already liquid there, it has to push that liquid out of the way, the only direction to go is up.
I can't remember how to prove it's a parabola but I don't think it's too difficult.

take a container of liquid and spin it... a parcel at any given radius (r) experiences (in the rotating frame) a force outward = mass x angular speed^2 x r and a vertical force of mg. Thus the surface curves such that the tangent to the surface makes an angle to the horizontal given by tan theta = angular speed^2 x r/g.

two questions:
-I think this is simple and I'm just missing it... how do I prove that the resultant shape is a parabola?

The curve of the shape of the liquid that you get has a tangent at every point whose slope is given by [tex]\large tan \theta[/tex] which is equal to [tex]\large dy/dx[/tex]. But [tex]\large tan \theta = \omega^{2} x/g[/tex]. On solving the equation you get

[tex]\large y= \omega^{2} x^{2}/2g[/tex] - the equation of a parabola.

Thank you -- I was looking at that earlier, but doesn't that imply that the angle from the vertex to any point equals the angle of the tangent line at a point? You replaced r with x which seems to me to say that but you started with the differential form. Thus, the right triangle composed of the tangent to the surface and dy and dx has the same angles as the triangle with the distance from the vertex to any point and the coordinates of the point?? I have trouble picturing that but then again my diagrams may be misleading me.

take a container of liquid and spin it... a parcel at any given radius (r) experiences (in the rotating frame) a force outward = mass x angular speed^2 x r and a vertical force of mg. Thus the surface curves such that the tangent to the surface makes an angle to the horizontal given by tan theta = angular speed^2 x r/g.

two questions:
-I think this is simple and I'm just missing it... how do I prove that the resultant shape is a parabola?

-what provides the force we see at work? surface tension? pressure gradient? Basically how
do we draw a free body diagram for the parcel of liquid in the Earth frame?
Thank you!

When the container is set rotating, (or part of the water itself, as by stirring), the motion is communicated to the whole of the water through viscous forces. Ultimately, when the system is in equilibrium in the rotating frame, there is a pressure gradient which balances the "weight" of the water. The total centrifugal force and weight of the whole water is ultimately balanced at the wall of the container, without which the water would fly apart.

Looked at from the static frame, there is a reaction force acting on any elementary volume of water, which is equal to the centripetal force. This reaction is similar to the case when a solid body is resting on a rotating container or the tension in the string when an object is being whirled around. As to what provides this force, it's always some kind of stress as in the case of the solid or the string. The water is under stress, one manifestation of which is the pressure.

then, would you call the force exerted on a parcel of water at the surface a normal force?

It has to be. If there's a tangential component, water would flow until only the normal compnent remains. Ask yourself why the surface of water on the earth is flat. It is also crucial in calculating the shape of the surface.

The force acting on an elementary mass dm at the surface is the w^2*x*dm acting along the x-axis outward and dm*g acting along the negative y-axis downward. If the resultant is normal to the curve, then the slope of the normal -dx/dy = (-dm*g)/(w^2*x*dm), from which the result follows.